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4 years ago

Find the difference. Round your answer to match the less precise measurement. 83 g – 1.8 g 81 g 82 g 80 g 81.2 g

Mathematics

1 answer:

Ivan4 years ago

6 0

Answer:

Option A.

Step-by-step explanation:

The given expression is

$83 g-1.8 g$

We need to find the difference round to the less precise measurement.

Taking out the common factor.

$(83-1.80)g$

$81.2g$

Precise measurement of given expression is 81.2g.

A measurement with more decimal places is more precise and vice versa.

Now, approx the answer to the the nearest whole measure.

$81.2g\approx 81g$

The less precise measurement of given expression is 81g.

Therefore, the correct option is A.

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Thirty students are eating lunch at 5 tables. Each table has the same number of students. How many students are sitting at each

kaheart [24]

30 ÷ 5
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3 years ago

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Max makes and sells posters. The function p(x)= -10x^2 +200x -250, graphed below, indicates how much profit he makes in a month

viktelen [127]

Here is our profit as a function of # of posters
p(x) =-10x² + 200x - 250
Here is our price per poster, as a function of the # of posters:
pr(x) = 20 - x
Since we want to find the optimum price and # of posters, let's plug our price function into our profit function, to find the optimum x, and then use that to find the optimum price:
p(x) = -10 (20-x)² + 200 (20 - x) - 250
p(x) = -10 (400 -40x + x²) + 4000 - 200x - 250
Take a look at our profit function. It is a normal trinomial square, with a negative sign on the squared term. This means the curve is a downward facing parabola, so our profit maximum will be the top of the curve.
By taking the derivative, we can find where p'(x) = 0 (where the slope of p(x) equals 0), to see where the top of profit function is.
p(x) = -4000 +400x -10x² + 4000 -200x -250
p'(x) = 400 - 20x -200
0 = 200 - 20x
20x = 200
x = 10
p'(x) = 0 at x=10. This is the peak of our profit function. To find the price per poster, plug x=10 into our price function:
price = 20 - x
price = 10
Now plug x=10 into our original profit function in order to find our maximum profit:
p(x)= -10x^2 +200x -250
p(x) = -10 (10)² +200 (10) - 250
p(x) = -1000 + 2000 - 250
p(x) = 750

Correct answer is C)

7 0

3 years ago

A meter stick casts a shadow 2.4 m long at the same time a flagpole casts a shadow 9.7 m long. How tall is the flagpole

Luda [366]

The first thing we are going to do is draw a diagram of the situation to help us to solve the situation.
We an draw tow similar triangle between the length of the shadows and the height of the stick and the pole.
Since meter stick has a length of 1 meter, the height of our smaller triangle is 1 meter. Let $h$ be height of of the pole. Since both triangles are similiar we can establish a proportion between their corresponding sides and solve for $h$ :
$\frac{h}{1} = \frac{9.7}{2.4}$
$h=\frac{9.7}{2.4}$
$h=4.04$

We can conclude that the pole is 4.04 meters tall.

8 0

3 years ago

1. What does it mean that Nikoleta is in the 10th percentile for length and weight? Is that a desirable statistic if she was in

VladimirAG [237]

Explanation:

1. It means that Nikoleta is longer and heavier than 10% of children her age.

2. We generally expect that a change from being in the 50th percentile to the 10th percentile signifies a condition that is interfering with normal growth. That is not usually considered desirable.

5 0

4 years ago

Read 2 more answers

4 Tan A/1-Tan^4=Tan2A + Sin2A

Eva8 [605]

tan(2A) + sin(2A) = sin(2A)/cos(2A) + sin(2A)

• rewrite tan = sin/cos

… = 1/cos(2A) (sin(2A) + sin(2A) cos(2A))

• expand the functions of 2A using the double angle identities

… = 2/(2 cos²(A) - 1) (sin(A) cos(A) + sin(A) cos(A) (cos²(A) - sin²(A)))

• factor out sin(A) cos(A)

… = 2 sin(A) cos(A)/(2 cos²(A) - 1) (1 + cos²(A) - sin²(A))

• simplify the last factor using the Pythagorean identity, 1 - sin²(A) = cos²(A)

… = 2 sin(A) cos(A)/(2 cos²(A) - 1) (2 cos²(A))

• rearrange terms in the product

… = 2 sin(A) cos(A) (2 cos²(A))/(2 cos²(A) - 1)

• combine the factors of 2 in the numerator to get 4, and divide through the rightmost product by cos²(A)

… = 4 sin(A) cos(A) / (2 - 1/cos²(A))

• rewrite cos = 1/sec, i.e. sec = 1/cos

… = 4 sin(A) cos(A) / (2 - sec²(A))

• divide through again by cos²(A)

… = (4 sin(A)/cos(A)) / (2/cos²(A) - sec²(A)/cos²(A))

• rewrite sin/cos = tan and 1/cos = sec

… = 4 tan(A) / (2 sec²(A) - sec⁴(A))

• factor out sec²(A) in the denominator

… = 4 tan(A) / (sec²(A) (2 - sec²(A)))

• rewrite using the Pythagorean identity, sec²(A) = 1 + tan²(A)

… = 4 tan(A) / ((1 + tan²(A)) (2 - (1 + tan²(A))))

• simplify

… = 4 tan(A) / ((1 + tan²(A)) (1 - tan²(A)))

• condense the denominator as the difference of squares

… = 4 tan(A) / (1 - tan⁴(A))

(Note that some of these steps are optional or can be done simultaneously)

7 0

3 years ago